2x^2+3x+4=
The equation you have given is a quadratic equation, which means it is a polynomial equation of degree 2. To solve this equation, we can use the quadratic formula or factorization method.
Let's solve it using the quadratic formula:
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / 2a, where 'a', 'b', and 'c' are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In your equation, we have:
a = 2
b = 3
c = 4
Substituting these values into the quadratic formula, we get:
x = (-3 ± √(3^2 - 4 * 2 * 4)) / (2 * 2)
Simplifying further,
x = (-3 ± √(9 - 32)) / 4
x = (-3 ± √(-23)) / 4
Since we have a negative value under the square root (√(-23)), we know that the solutions will involve complex numbers. Specifically, we have an imaginary number because we cannot take the square root of a negative number in the real number system.
Therefore, the solutions to the equation 2x^2 + 3x + 4 = 0 are:
x = (-3 + √(-23)i) / 4
and
x = (-3 - √(-23)i) / 4
The + and - in the solutions indicate that there are two complex solutions.